By Radhika, T.S.L

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Are in terms of a1. We know that a0 ↑ 0 . Let us find the value of a1. 15), which gives a1 = 0 . 30 A p p r ox im at e A n a ly ti c a l M e t h o d s Now, using the recurrence relation given previously in this example, we obtain a0 3! 4 5! a2 = − and so on. Thus, the first Frobenius solution is y1 ( x ) = a0 x 1/2 1 − x2 x4 + − . 2! 5! Now, y 2 = a 0* x 1/2 1 − x2 x4 + − 2! 5! ∫ 1 x 1/2 x2 x4 1− + − 2! 5! 2 e − 1 ∫ x dx After straightforward but lengthy calculations, we obtain y 2 = a 0* x −1/2 1 − x2 x4 + − 2!

Otherwise, x0 is termed an IRSP. For instance, (i) x = 0 is an RSP of y − cos x y = 0. x Note that P (x ) = − cos x 1 x x3 =− + − + x x 2 24 24 A p p r ox im at e A n a ly ti c a l M e t h o d s (ii) x = 1 is an IRSP of y − 1 y = 0. 11) if ( x − x0 ) P ( x ) is analytic at x0 . At RSPs, we look for solutions of the form ∞ y= ∑ a (x − x ) n n=0 0 n +m , a0 ≠ 0 because if a0 = 0 , then some positive integral power of x can be factored out of the power series part and can be combined with ( x − x0 )n .

Van Ekeren. A Treatise on the Hydrogen Bomb. Hamilton, NZ: University of Waikoto, 2008. P. Koscik and A. Kopinska. Application of the Frobenius method to the Schrodinger equation for a spherically symmetric potential: An harmonic oscillator. Journal of Physics A: Mathematics and General, Vol. 38, pp. 7743–7755, 2005. R. Ballarini and P. Villaggio. Frobenius method for curved cracks. International Journal of Fracture, Vol. 139, pp. 59–69, 2006. M. Apostol. Mathematical Analysis. 2nd edition. Boston: Addison-Wesley, 1974.

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